An arrow is shot and follows the path described by the equation h(t)=-5t^2+29t+2, where h is the hight of the arrow in meters and t is the time in seconds. a) how long is the arrow in the air? b)what is the maximum height of the arrow? c) for how long was the arrow at least 3.0m off the ground?

A) when h(t)=0 the arrow is on the ground, so 5t²-29t-2=0. Using the quadratic formula we get t=(2.9±√881/10).
This gives us two values for t. The difference between the two values is the length of time the arrow is in the air=√881/5=5.94 seconds.
b) the maximum height is when the speed of the arrow is zero, which is given by h’(t)=-10t+29=0, so t=2.9 seconds.
h(2.9)=44.05m.
c) -5t²+29t+2≥3, 5t²-29t+1≤0. Using the quadratic formula again, solve the equality before the inequality:
t=2.9±√821/10 seconds. The graph is an upright U-shaped parabola cutting the t axis at 0.0347 and 5.7653. Therefore between these times the graph is below the t axis where h(t)≥3, meaning that the arrow is at least 3m off the ground.